# Question Video: Finding the Limit of a Composition of Rational Functions at Infinity Mathematics • 12th Grade

Find lim_(π₯ β β) ((3π₯/(8π₯ + 4)) β (7π₯Β²/(2π₯ + 7)Β²)).

07:49

### Video Transcript

Find the limit as π₯ approaches infinity of the function three π₯ over eight π₯ plus four minus seven π₯ squared divided by two π₯ plus seven squared.

First, we write out the whole limit. We might be tempted to first try simplifying our expression to be a single fraction. But we will see that this is a more-difficult-than-necessary approach. To do this, we want the denominators of the fraction to be equal. So we would use cross multiplication. We multiply the numerator three π₯ by the denominator two π₯ plus seven all squared. We then subtract the numerator seven π₯ squared multiplied by eight π₯ plus four. And our denominator is found by multiplying the two denominators together. So eight π₯ plus four multiplied by two π₯ plus seven squared.

At this point, we would want to use our rule for finding limits of rational functions. Which says that we should divide the numerator and the denominator by the highest power of π₯ found in the denominator. So we can ask what is the highest power of π₯ found in the denominator? We see that the highest power of π₯ found in the denominator will be obtained when we multiply eight π₯ and two π₯ squared. So the highest power of π₯ will be π₯ cubed.

We would then divide the numerator and the denominator by π₯ cubed. While this would work, we can see that this will be cumbersome due to the amount of algebraic terms present. Instead, we could start again. Except this time, we will start by using that the limit of the difference between two functions is equal to the difference of the limits of these two functions. This gives us the limit of three π₯ over eight π₯ plus four as π₯ approaches infinity. Minus the limit of seven π₯ squared over two π₯ plus seven squared as π₯ approaches infinity.

Now we want to use our rule of dividing by the highest power of π₯ found in the denominator to help us evaluate both of the limits of these two rational functions. If we start from the first limit, we see that the denominator is linear. So the highest power of π₯ is just π₯. So we will divide both the numerator and the denominator of this limit by π₯. In the second limit, we have a denominator of two π₯ plus seven squared. The highest power of π₯ here will be when we square two π₯ to give us four π₯ squared. So we have the highest power in the denominator of our second limit is equal to π₯ squared. Therefore, we will divide the numerator and the denominator of this limit by π₯ squared.

We see that the numerator of our first limit will be three π₯ over π₯. And we can cancel this shared factor of π₯ to give us the value of three. Therefore, we can just replace the numerator in our expression with just three. The denominator of our first limit will be eight π₯ plus four over π₯. We can split this into the sum of eight π₯ over π₯ plus four over π₯, which simplifies to give us eight plus four over π₯. So we can replace the denominator in our first limit with eight plus four over π₯.

We now move on to our second limit. The numerator in our second limit will be seven π₯ squared divided by π₯ squared. Which we can simplify by canceling the shared factor of π₯ squared, to just give us seven. So we can replace the numerator in our second limit with just the value of seven. The denominator in our second limit is two π₯ plus seven squared over π₯ squared. We can then use the fact that π plus π all squared divided by π squared is equal to π plus π over π all squared. To rewrite our denominator as two π₯ plus seven over π₯ all squared. Just as we did before, we can split this into a sum, giving us two π₯ over π₯ plus seven over π₯ all squared. And then, we can cancel the shared factor of π₯, giving us two plus seven over π₯ all squared. And so we can change the denominator in our second limit to be two plus seven over π₯ all squared.

Now we can use the quotient rule for limits. Which says that the limit of a quotient of two functions is equal to the quotient of the limit of these two functions. Though we can use this to rewrite our first limit as the quotient of two limits. This gives us the limit of three divided by the limit of eight plus four over π₯. And we can also rewrite our second limit as the quotient of two limits. This gives us the limit of seven as π₯ approaches infinity divided by the limit of two plus seven over π₯ squared as π₯ approaches infinity. Now, weβre going to use the fact that, for any constant π, the limit of π as π₯ approaches π is just equal to π. So the limit of three in our numerator is just equal to three. And the limit of seven in our numerator is just equal to seven.

We also know that the limit of the sum of two functions is equal to the sum of the limit of those two functions. You can apply this to the limit in the denominator of our first fraction to get the limit of eight as π₯ approaches infinity plus the limit of four over π₯ as π₯ approaches infinity. Again, the limit of the constant eight as π₯ approaches infinity is just eight. We also know that, for any constant π, the limit of π multiplied by π of π₯ as π₯ approaches π is equal to π multiplied by the limit of π π₯ as π₯ approaches π. We can use this to see that the limit of four over π₯ is equal to four multiplied by the limit of one over π₯.

The next thing we want to use is that the limit of the reciprocal function as π₯ approaches infinity is equal to zero. So we can evaluate the limit of the reciprocal function in our denominator to just be zero. Which means we can simplify the first denominator to just be equal to eight. The last limit rule that weβre going to use is that the limit of a power is equal to the power of the limit. We can use this to see that the limit of two plus seven over π₯ squared is equal to the square of the limit of two plus seven over π₯. We can then use the fact that the limit of a sum is equal to the sum of the limits to split the limit in our denominator. This gives us the limit of two as π₯ approaches infinity plus the limit of seven over π₯ as π₯ approaches infinity all squared.

We know that the limit of two as π₯ approaches infinity is just equal to two. We can then take the constant of seven out of our limit. Giving us seven multiplied by the limit of the reciprocal function as π₯ tends to infinity. And we also know that we can evaluate the limit of the reciprocal function as π₯ approaches infinity to just be zero. This gives us a denominator of two plus seven multiplied by zero all squared, which we can evaluate to be four. This gives us that the limit in our question is equal to three π₯ minus seven over four, which we can evaluate to be negative 11 divided by eight.

Giving us that the limit of the function three π₯ divided by eight π₯ plus four minus seven π₯ squared over two π₯ plus seven squared is equal to negative 11 divided by eight.